JEE Mathematics 2025 Question with Solution

Given: A relation $R$ on $\left[0, \frac{\pi}{2}\right]$ is defined by

Find: Whether $R$ is reflexive, symmetric, transitive, and hence whether it is an equivalence relation.

Using the identity

we compare the given condition with this standard trigonometric result.

Reflexivity: For $xRx$, we need

Since

we get

Therefore, $R$ is reflexive.

Symmetry: If $xRy$, then

For $yRx$, we need

Using

and

we obtain

Thus, $R$ is symmetric.

Transitivity: Assume $xRy$ and $yRz$. Then

and

So,

and

Using $\sec^2 y = 1 + \tan^2 y$, we get

which gives the required simplification in the solution, and hence

Therefore, $xRz$ and $R$ is transitive.

Since $R$ is reflexive, symmetric, and transitive, $R$ is an equivalence relation. Therefore, the correct option is D.

From

we may rewrite

But

so the condition becomes

Thus the relation is effectively comparing the same trigonometric quantity for $x$ and $y$.

Any relation of the form “the value of a function at $x$ equals the value of the same function at $y$” is reflexive, symmetric, and transitive. Hence $R$ is an equivalence relation, so the correct option is D.